Question

Are the vectors v1=(1,1,2,4), v2=(2,-1,-5,2),v3=(1,-1,-4,0), v4=(2,1,1,6) independent in R^4. Please find the basis spanned by these vectors.

Answer 1

Is there a nontrivial linear combination of them which sums to 0 in R^4? Let's find out by supposing a, b, c, and d are constants such that a(v1)+...+d(v4)=0. Then we get the system of equations \[\begin{align*}a+2b+c+2d&=0\\a-b-c+d&=0\\2a-5b-4c+d&=0\\4a+2b+6d&=0\end{align*}\]

You can use elimination to solve this system. If you can come up with non-zero values of a, b, c, and d, then those values give a linear combination which exhibits dependence. If there is a dependence, remove one of the vectors and see if the remaining three are independent.

I have to leave where I'm working right now, but when I get home I'll type more detail. In the mean time, let me know what you come up with.

Answer 2

a very easy way is use from determinant , if determinant of |v1,..,v4|is not equal to zero then 4 vector are independent and they can used as a basis for R4

Answer 3

One drawback of that method is that it gives you no sense of which vectors to remove. For instance, suppose that v1 = v2 but that {v2, v3, v4} is independent. After computing | (v1 v2 v3 v4) | = 0 you will be unsure which vector to remove when, clearly, you only want to remove v1 or v2. The case is similar when v1 is a multiple of v2, or also if v3 is a linear combination of v1 and v2, for instance.

The method is pretty good for when dim(V)<4 because it's relatively easy to figure out whether some two vectors are multiples of each other, and in the case when one isn't a multiple of the other but v3 = av1 + bv2, then you can delete any one of the vectors and obtain a basis. But for dim(V) >/= 4, the determinant method suffers.

Answer 4

Or at least suffers when they are dependent.

Answer 5

@AddemF If you are looking for a way to know which of the infringing vectors to remove, perform a full row reduction. After, all free columns are the same columns which represent the vectors which, if removed, will create a fully LI basis from the pivot columns. This works b/c elementary row ops cannot swap columns.

Answer 6

Sure, row reduction will do the trick--I was only responding to the use of the determinant.

Answer 7

@AddemF I'm sure you knew about this, I was just posting it for @sadananda42 to see b/c you left that part out in your explanation.